[AMath 383] Quizzes
Dear class members,
I have updated the grade record of quizzes. Please remember that you
are responsible for checking on quiz concepts that you haven't passed.
There have been 9 quiz concepts so far. Tomorrow will introduce the
10th. If your online grade doesn't show all nine, then you have haven't
taken it yet. The concepts associated with the quizzes are listed on
the webpage.
The bifurcation diagram quizzes seems to be giving many of you a strong
challenge. Some who have passed previously seem to demonstrate that
they didn't really understand. (Policy, however, dictates that you
still receive credit.) Monday, we had a quiz where the diagram was
already drawn, and you only needed to determine the stability of the
different branches.
You were given the growth rate dx/dt = f(x) as a function.
Major Mistake #1: Trying to use f'(x) at an arbitrary point.
Recall that f(x) gives the rate of change at a point x. If f>0, x is
increasing. If f<0, x is decreasing. The function f'(x) doesn't say
this at all; it says something about how fast the rate of change (f) is
changing once you move.
Question: So when do we use f'(x)?
Answer: If x=a is an equilibrium, then we know that f(a)=0. So we don't
know whether nearby points are increasing or decreasing. This is when
we use f'(x).
If f'(a)>0, then f(x)>0 for x>a (i.e., x increasing). But f(x)<0 for
x<a (i.e., x decreasing). Thus x=a is an unstable equilibrium.
If f'(a)<0, then f(x)<0 for x>a (i.e., x decreasing). But f(x)>0 for
x<a (i.e., x increasing). Thus this time, x=a is a stable equilibrium.
Question: How do we deal with a parameter when using f'(x)?
Answer: So we assume that an equilibrium occurs for x=a(mu), where mu
is the parameter value and a(mu) says how the equilibrium depends on
mu. Then we can test the stability of the equilibrium a(mu) using
f'(a(mu)).
Example: f(x) = x(x-mu) = x^2 - mu x
f'(x) = 2x - mu
Equilibria occur when f(x)=0, i.e., x=0 or x=mu.
Let a1(mu)=0 describe the first branch
and let a2(mu) = mu describe the second branch
f'(a1(mu)) = f'(0) = -mu
So the branch a1 is stable if mu>0 (when f'(a1)<0)
and unstable when mu<0 (when f'(a1)>0)
f'(a2(mu)) = f'(mu) = 2(mu) - mu = mu
So the branch a2 is unstable if mu>0 (when f'(a2)>0)
and stable when mu<0 (when f'(a2)<0).
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Aside: At a bifurcation point, multiple branches intersect, which has a
consequence of f'(a)=0. (x=a is a multiple root of f(x) at this value
of mu.) So stability must be checked away from the bifurcation.
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Alternative approach:
Instead of looking right at the equilibrium, where you need to compute
f'(x), consider a point that is off the branches of the bifurcation
diagram. Choose a parameter value (mu) and a value (x), and just plug
these into f(x).
(NOT f'(x)!!!!)
Then if f(x)>0, you know that the solution is increasing. If f(x)<0,
you know that the solution is decreasing. Any point that is in the same
region of the bifurcation diagram without needing to cross a branch of
equilibria will have growth in the same direction. Ideally, you should
check a point in each separated region, but you can avoid this often by
just checking whether the sign of the growth rate changes if you cross
a branch.
I hope this helps.
But tomorrow's quiz will actually be on nullclines. (But the same
comments apply in determining what direction the solutions will move in
the regions separated by nullclines.)
Sincerely,
D. Brian Walton
AMath 383 Instructor
Guggenheim 408C
(206) 685-9298